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3 years ago

Ryan surveyed parents. He found 3 out of every 5 parents vote. 450 parents eligible. How many parents will vote in the election?

Mathematics

1 answer:

Dafna1 [17]3 years ago

4 0

The answer is 150 parents. Because 450/3=150

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the metro theater has 20 rows of seats with 18 seats I each row. Ticket cost $5. the theater's income in dollars if all seat are

Mumz [18]

500 IS THE ANSWER XD

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3 years ago

1 -12.166 + (-1.54)

castortr0y [4]

Answer:

-12.706

Step-by-step explanation:

First we distribute: 1 - 12.166 - 1.54

Then we subtract: -12.706

8 0

3 years ago

Duane and mick are saving their money to build a tree house. Duane adds $5 to their piggy bank every other week. Mick adds $2 er

Furkat [3]

10 weeks
Duane adds $5 every other week so that's 5 weeks out of 10. 5 x 5 = 25
Mick adds $2 every week so,
2 x 10 = 20

25 + 20 = 45

8 0

3 years ago

Will give brainliest A ? B? C ?

dolphi86 [110]

Answer:

A: 180° B: 360° C: 540°

Step-by-step explanation:

A: The measures of interior angles of a triangle sum to 180°. Since figure ABC is a triangle, the interior angles of figure ABC sum to 180°.

B: We can see the quadrilateral ACDE is a trapezoid. If you can see, the trapezoid can be split into two triangles if you connect points C and E. As U mentioned before, the measures of interior angles of a triangle sum to 180°. Since we know figure ACDE can be split into two triangles we have to do: 180x2 which is 360°.

C: Now that we know the Pentagon ABCDE can be split into 3 triangles(1 from figure ABC and 2 from figure ACDE), we have to multiply 180 by 3 because like I mentioned before, the measures of interior angles of a triangle sum to 180°. So, 180x3=540°.

5 0

3 years ago

Read 2 more answers

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0

3 years ago